Let $\textbf{$\gamma$} (t)$ be a regular plane curve and let $\lambda$ be a constant. The parallel curve $\textbf{$\gamma$}^{\lambda}$ of $\textbf{$\gamma$}$ is defined by $\textbf{$\gamma$}^{\lambda}(t) = \textbf{$\gamma$} (t) + \lambda \textbf{n}_s(t)$.
Show that, if $\lambda \kappa_s(t) \neq 1$ for all values of $t$, then $\textbf{$\gamma$}^{\lambda}$ is a regular curve and that its signed curvature is $\frac{\kappa_s}{|1 − \lambda \kappa_s|}$.
$$$$
I have shown that if $\lambda \kappa_s(t) \neq 1$ for all values of $t$, then $\textbf{$\gamma$}^{\lambda}$ is a regular curve.
I am facing some difficulties at showing that the signed curvature is $\frac{\kappa_s}{|1 − \lambda \kappa_s|}$.
I have done the following for this part:
The signed curvature $\kappa_s^{\lambda}$ is equal to $$\kappa_s^{\lambda}=\textbf{$\gamma$}^{\lambda ''}\textbf{n}_s^{\lambda}$$
$\textbf{n}_s$ is the signed unit perpendicular vector to $\textbf{$\gamma$}$. Since $\textbf{$\gamma$}^{\lambda}$ is parallel to $\textbf{$\gamma$}$ and $\textbf{n}_s^{\lambda}$ is the signed unit perpendicular vector to $\textbf{$\gamma$}^{\lambda}$, we have that $\textbf{n}_s | | \textbf{n}_s^{\lambda}$, i.e., $ \textbf{n}_s^{\lambda}=a\textbf{n}_s$.
We have that $\textbf{t}=\frac{\textbf{$\gamma$}'(t)}{\|\textbf{$\gamma$} '(t)\|}=\frac{\textbf{$\gamma$}'(t)}{s'(t)} \Rightarrow \textbf{$\gamma$}'(t)=s'(t)\textbf{t}$.
So, we have $$\textbf{$\gamma$}^{\lambda '}(t) = \dot s\mathbf t (1- \lambda \kappa_s ) \Rightarrow \textbf{$\gamma$}^{\lambda '}(t) = \dot s\mathbf t - \lambda \kappa_s \dot s\mathbf t \Rightarrow \textbf{$\gamma$}^{\lambda '}(t) = \dot s\mathbf t + \lambda \textbf{n}_s'$$
Differentiating the relation $$\textbf{$\gamma$}^{\lambda '}(t) = \dot s\mathbf t + \lambda \textbf{n}_s'$$ we get $$\textbf{$\gamma$}^{\lambda ''}(t) = \ddot s\mathbf t +\dot s\mathbf {\dot t} + \lambda \textbf{n}_s''$$
Then we have $$\kappa_s^{\lambda}=\textbf{$\gamma$}^{\lambda ''}\textbf{n}_s^{\lambda} \Rightarrow \kappa_s^{\lambda}=\left (\ddot s\mathbf t +\dot s\mathbf {\dot t} + \lambda \textbf{n}_s''\right ) a\textbf{n}_s$$
Is this correct so far? How do we get from here the desired result?
Start with the definition of the parallel curve $\gamma^{\lambda}$ and take it's derivative with respect to time. Then,
\begin{align} \textbf{$\gamma$}^{\lambda'}(t) &= \textbf{$\gamma'$} (t) + \lambda \textbf{n}_s'(t) \\ \textbf{$\gamma$}^{\lambda'}(t) &= \dot s(t)\textbf{t} - \lambda \kappa_s \dot s(t)\textbf{t} \\ \textbf{$\gamma$}^{\lambda'}(t) &= \left(1 - \lambda \kappa_s\right)\dot s(t)\textbf{t} \\ \dot s^{\lambda}(t)\textbf{t}^{\lambda} &= \left(1 - \lambda \kappa_s\right)\dot s(t)\textbf{t} \end{align}
Since $\textbf{t}^{\lambda} \parallel \textbf{t}$, \begin{equation} \dot s^{\lambda}(t) = \left(1 - \lambda \kappa_s\right)\dot s(t) \\ \frac {ds^{\lambda}(t)}{dt} = \left(1 - \lambda \kappa_s\right)\frac {ds(t)}{dt} \end{equation}
Arc length for the parallel curve has to be distinguished from arc length for the original curve. Same with unit tangent. Then,
\begin{align} \textbf{t} &= \frac{\textbf{$\gamma$}'(t)}{s'(t)} \\ \textbf{t}^{\lambda} &= \frac{\textbf{$\gamma$}^{\lambda'}(t)}{s^{\lambda'}(t)} = a \textbf{t} \\ \textbf{n}_s^{\lambda} &= a\textbf{n}_s \end{align}
Then the signed curvature $\kappa_s^{\lambda}$ is
\begin{align} \kappa_s^{\lambda}\textbf{n}_s^{\lambda}=\frac{d\textbf{t}^{\lambda}}{ds^{\lambda}(t)} = \dfrac {\frac {d\textbf{t}^{\lambda}}{dt}}{\frac {ds^{\lambda}(t)}{dt}} \end{align}
By substitution,
\begin{align} \kappa_s^{\lambda}\textbf{n}_s^{\lambda} &= \dfrac {a\frac {d\textbf{t}}{dt}}{\left(1 - \lambda \kappa_s\right)\frac {ds(t)}{dt}} \\ \kappa_s^{\lambda}\textbf{n}_s^{\lambda} &= a \left(1 - \lambda \kappa_s \right)^{-1} \frac {d\textbf{t}} {ds(t)} \end{align}
By the definition of signed curvature, $\kappa_s \textbf{n}_s = \frac {d\textbf{t}} {ds(t)}$, so by substitution
\begin{align} \kappa_s^{\lambda}\textbf{n}_s^{\lambda} &= a \kappa_s \left(1 - \lambda \kappa_s \right)^{-1} \textbf{n}_s \end{align}
Since $\textbf{n}_s^{\lambda} = a\textbf{n}_s$, or $\frac{1}{a}\textbf{n}_s^{\lambda} = \textbf{n}_s$,
\begin{align} \kappa_s^{\lambda}\textbf{n}_s^{\lambda} &= \frac {a}{a} \kappa_s \left(1 - \lambda \kappa_s \right)^{-1} \textbf{n}_s^{\lambda} \\ \kappa_s^{\lambda}\textbf{n}_s^{\lambda} &= \kappa_s \left(1 - \lambda \kappa_s \right)^{-1} \textbf{n}_s^{\lambda} \end{align}
Thus,
\begin{equation} \kappa_s^{\lambda} = \dfrac{\kappa_s} {\left(1 - \lambda \kappa_s \right)} \end{equation}
I'm not sure about the absolute value...